h =.1 = (-.114942587)
h =.01 = (-.1114827202)
h = -.01 = (-.1107419712)
h = -.1 = (-.1075268817)
I understood how to get these values but then it asked:
If the slope of the tangent line to the graph of F(x) at x = 0 was -1/n^2 for some integer n, what would you expect n to be?
n = ?
This is the part I do not understand, could someone please explain this to me.
Thank You,
Keith Stevens

March 7th 2007, 08:45 AM

Jhevon

Quote:

Originally Posted by kcsteven

Let f(x) = 1/x-3. Calculate the difference quotient F(0+h)-f(0)/h for

h =.1 = (-.114942587)
h =.01 = (-.1114827202)
h = -.01 = (-.1107419712)
h = -.1 = (-.1075268817)
I understood how to get these values but then it asked:
If the slope of the tangent line to the graph of F(x) at x = 0 was -1/n^2 for some integer n, what would you expect n to be?
n = ?
This is the part I do not understand, could someone please explain this to me.
Thank You,
Keith Stevens

i didnt check this, but you got the answers to be roughly -0.1 right. so we want n so that -1/n^2 = -0.1. so idealy we want n^2 = 10. since -1/10 = -0.1. but that would mean n = sqrt(10) which is not an integer. so the closest we can get is say, n^2 = 9. so -1/n^2 = -1/9 = -0.111111111... so then n=3

March 7th 2007, 09:25 AM

qbkr21

Quote:

Originally Posted by kcsteven

Let f(x) = 1/x-3. Calculate the difference quotient F(0+h)-f(0)/h for

h =.1 = (-.114942587)
h =.01 = (-.1114827202)
h = -.01 = (-.1107419712)
h = -.1 = (-.1075268817)
I understood how to get these values but then it asked:
If the slope of the tangent line to the graph of F(x) at x = 0 was -1/n^2 for some integer n, what would you expect n to be?
n = ?
This is the part I do not understand, could someone please explain this to me.
Thank You,
Keith Stevens

Hi Keith it's Andrew. Jhevon was right but I posted some steps that might give you more of a visual perspective.